Number of Longest Increasing Subsequences

Last Updated : 09 Dec, 2020

Given an array arr[] of size N, the task is to count the number of longest increasing subsequences present in the given array.

Examples:

Input: arr[] = {2, 2, 2, 2, 2}
Output: 5
Explanation: The length of the longest increasing subsequence is 1, i.e. {2}. Therefore, count of longest increasing subsequences of length 1 is 5.

Input: arr[] = {1, 3, 5, 4, 7}
Output: 2
Explanation: The length of the longest increasing subsequence is 4, and there are 2 longest increasing subsequences, i.e. {1, 3, 4, 7} and {1, 3, 5, 7}

Naive Approach: The simplest approach is to generate all possible subsequences present in the given array arr[] and count the increasing subsequences of maximum length. Print the count after checking all subsequences.

Time Complexity: O(N*2^N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming as the above problem has overlapping subproblems that need to be calculated more than once, and to reduce that calculation use tabulation or memoization. Follow the steps below to solve the problem:

Initialize two arrays dp_l[] and dp_c[] to store the length of the longest increasing subsequences and the count of the longest increasing subsequence at each index respectively.
Initialize a variable count with 0 to store the number of the longest increasing subsequence.
Iterate over the range [1, N – 1] using the variable i:
- Iterate over the range [0, i – 1] using the variable j:
  - If arr[j] >= arr[i] and dp_l[j]+1 > dp_l[i], then update dp_l[i] as dp_l[j] + 1 and dp_c[i] as dp_c[j].
  - Else if (dp_l[j] + 1) is the same as dp_l[i], then update dp_c[i] as dp_c[i] + dp_c[j].
Find the maximum element in the array dp_l[] and store it in a variable lis that will give the length of LIS.
Traverse the arrays dp_l[] and dp_c[] simultaneously and if at any index idx, dp_l[idx] is the same as lis then increment the count by dp_c[idx].
After the above steps, print the value of count as the resultant count.

Below is the implementation of the above approach:

C++

// C++ program for the 
// above approach
#include <bits/stdc++.h>
using namespace std;
 
//Function to count the number 
// of LIS in the array nums[]
int findNumberOfLIS(vector<int> nums)
{
  //Base Case
  if (nums.size() == 0)
    return 0;
 
  int n = nums.size();
 
  //Initialize dp_l array with 
  // 1s
  vector<int> dp_l(n, 1);
 
  //Initialize dp_c array with 
  // 1s
  vector<int> dp_c(n, 1);
 
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; j < i; j++)
    {
      //If current element is 
      // smaller
      if (nums[i] <= nums[j])
        continue;
 
      if  (dp_l[j] + 1 > dp_l[i])
      {
        dp_l[i] = dp_l[j] + 1;
        dp_c[i] = dp_c[j];
      }
      else if (dp_l[j] + 1 == dp_l[i])
        dp_c[i] += dp_c[j];
    }
  }
 
  //Store the maximum element 
  // from dp_l
  int max_length = 0;
 
  for (int i : dp_l)
    max_length = max(i,max_length);
 
  //Stores the count of LIS
  int count = 0;
 
  //Traverse dp_l and dp_c 
  // simultaneously
  for(int i = 0; i < n; i++)
  {
    //Update the count
    if (dp_l[i] == max_length)
      count += dp_c[i];
  }
   
  //Return the count of LIS
  return count;
}
 
//Driver code
int main()
{
  //Given array arr[]
  vector<int> arr = {1, 3, 5, 4, 7};
 
  //Function Call
  cout << findNumberOfLIS(arr) << endl;
}
 
// This code is contributed by Mohit Kumar 29

Java

// Java program for the 
// above approach
import java.util.*;
 
class GFG{
 
// Function to count the number 
// of LIS in the array nums[]
static int findNumberOfLIS(int[] nums)
{
   
  // Base Case
  if (nums.length == 0)
      return 0;
   
  int n = nums.length;
   
  // Initialize dp_l array with 
  // 1s
  int[] dp_l = new int[n];
  Arrays.fill(dp_l, 1);
 
  // Initialize dp_c array with 
  // 1s
  int[] dp_c = new int[n];
  Arrays.fill(dp_c, 1);
 
  for(int i = 0; i < n; i++)
  {
    for(int j = 0; j < i; j++)
    {
       
      // If current element is 
      // smaller
      if (nums[i] <= nums[j])
        continue;
 
      if (dp_l[j] + 1 > dp_l[i])
      {
        dp_l[i] = dp_l[j] + 1;
        dp_c[i] = dp_c[j];
      }
      else if (dp_l[j] + 1 == dp_l[i])
        dp_c[i] += dp_c[j];
    }
  }
 
  // Store the maximum element 
  // from dp_l
  int max_length = 0;
 
  for(int i : dp_l)
    max_length = Math.max(i, max_length);
 
  // Stores the count of LIS
  int count = 0;
 
  // Traverse dp_l and dp_c 
  // simultaneously
  for(int i = 0; i < n; i++)
  {
     
    // Update the count
    if (dp_l[i] == max_length)
      count += dp_c[i];
  }
   
  // Return the count of LIS
  return count;
}
 
// Driver code
public static void main(String[] args)
{
   
  // Given array arr[]
  int[] arr = { 1, 3, 5, 4, 7 };
 
  // Function Call
  System.out.print(findNumberOfLIS(arr) + "\n");
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach
 
# Function to count the number of LIS
# in the array nums[]
def findNumberOfLIS(nums):
 
    # Base Case
    if not nums:
        return 0
 
    n = len(nums)
 
    # Initialize dp_l array with 1s
    dp_l = [1] * n
 
    # Initialize dp_c array with 1s
    dp_c = [1] * n
 
    for i, num in enumerate(nums):
        for j in range(i):
 
            # If current element is smaller
            if nums[i] <= nums[j]:
                continue
 
            # Otherwise
            if dp_l[j] + 1 > dp_l[i]:
                dp_l[i] = dp_l[j] + 1
                dp_c[i] = dp_c[j]
 
            elif dp_l[j] + 1 == dp_l[i]:
                dp_c[i] += dp_c[j]
 
    # Store the maximum element from dp_l
    max_length = max(x for x in dp_l)
 
    # Stores the count of LIS
    count = 0
 
    # Traverse dp_l and dp_c simultaneously
    for l, c in zip(dp_l, dp_c):
 
        # Update the count
        if l == max_length:
            count += c
 
    # Return the count of LIS
    return count
 
# Driver Code
 
# Given array arr[]
arr = [1, 3, 5, 4, 7]
 
# Function Call
print(findNumberOfLIS(arr))

C#

// C# program to implement 
// the above approach 
using System;
using System.Collections.Generic;
class GFG {
     
    // Function to count the number 
    // of LIS in the array nums[]
    static int findNumberOfLIS(int[] nums)
    {
        
      // Base Case
      if (nums.Length == 0)
          return 0;
        
      int n = nums.Length;
        
      // Initialize dp_l array with 
      // 1s
      int[] dp_l = new int[n];
      Array.Fill(dp_l, 1);
      
      // Initialize dp_c array with 
      // 1s
      int[] dp_c = new int[n];
      Array.Fill(dp_c, 1);
      
      for(int i = 0; i < n; i++)
      {
        for(int j = 0; j < i; j++)
        {
            
          // If current element is 
          // smaller
          if (nums[i] <= nums[j])
            continue;
      
          if (dp_l[j] + 1 > dp_l[i])
          {
            dp_l[i] = dp_l[j] + 1;
            dp_c[i] = dp_c[j];
          }
          else if (dp_l[j] + 1 == dp_l[i])
            dp_c[i] += dp_c[j];
        }
      }
      
      // Store the maximum element 
      // from dp_l
      int max_length = 0;
      
      foreach(int i in dp_l)
        max_length = Math.Max(i, max_length);
      
      // Stores the count of LIS
      int count = 0;
      
      // Traverse dp_l and dp_c 
      // simultaneously
      for(int i = 0; i < n; i++)
      {
          
        // Update the count
        if (dp_l[i] == max_length)
          count += dp_c[i];
      }
        
      // Return the count of LIS
      return count;
    }
 
  // Driver code
  static void Main() {
       
      // Given array arr[]
      int[] arr = { 1, 3, 5, 4, 7 };
      
      // Function Call
      Console.WriteLine(findNumberOfLIS(arr));
  }
}
 
// This code is contributed by divyeshrabadiya07

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N)

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